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Misha Larkins [42]
3 years ago
11

Snowy's Snow Cones has a special bubble gum snow cone on sale. The cone is a regular snow cone that has a spherical piece of bub

ble gum nested at the bottom of the cone. The radius of the snow cone is 6 inches, and the height of the cone is 10 inches. If the diameter of the bubble gum is 1.5 inches, which of the following can be used to calculate the volume of the cone that can be filled with flavored ice? 1 over 3(3.14)(102)(6) − 4 over 3(3.14)(1.53) 1 over 3(3.14)(62)(10) − 4 over 3(3.14)(1.53) 1 over 3(3.14)(102)(6) − 4 over 3(3.14)(0.753) 1 over 3(3.14)(62)(10) − 4 over 3(3.14)(0.753)
Mathematics
2 answers:
LekaFEV [45]3 years ago
6 0
Hello again!! (go to your messages if you don't remember me)

First, calculate the volume of the empty cone.
V₁ = (1/3)π*(6 in)²*(10 in) = 120π in³

Calculate the volume of the sphere.
V₂ = (4/3)π*(1.5 in)³ = 4.5π in³

The volume that can be filled with flavored ice is 
V = V₁ - V₂ = 115.5π in³ = 362.85 in³

Answer:
The volume is 115.5π in³ or 362.9 in³ (nearest tenth)

Hope this helps!! Let me know if you have ANY questions.
posledela3 years ago
3 0

Answer:

The radius of the snow cone is 6 inches, and the height of the cone is 10 inches.

The diameter of the bubble gum is 1.5 inches.

The volume of cone is given by = \frac{\pi r^{2} h}{3}

So, volume of cone is =  \frac{3.14*6*6* 10}{3}= 376.80 cubic inches

And as the bubble gum is a sphere of diameter 1.5 inches.

So, radius is \frac{1.5}{2}= 0.75 inches

Volume of sphere is given by = \frac{4}{3} \pi r^{3}

Volume becomes: \frac{4}{3} *3.14*0.75*0.75*0.75

= 1.766 cubic inches

So, volume of the cube that can be filled with flavored ice will be =

376.80-1.766=375.03 cubic inches.

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Answer:

440

Step-by-step explanation:

1.5 hours he 90 minutes

660/ 90=7.333

7.333*60 (which is how many minutes in an hour = 440 miles

6 0
3 years ago
7/7q+21= x /5q^2-45 then x=?​
anzhelika [568]

Answer:

x = 5q - 15

Step-by-step explanation:

\frac{7}{7q+21}=\frac{x}{5q^{2}-45}\\\\\frac{7}{7(q+3)}=\frac{x}{5 (q^{2} -9)}\\\\frac{1}{q+3}=\frac{x}{5*(q^{2}-3^{2})}\\\\\frac{1}{q+3}=\frac{x}{5(q+3)(q-3)}\\\\\frac{1}{q+3}*5*(q+3)(q-3)=x\\\\5(q-3)=x\\\\x= 5q-15

3 0
3 years ago
Log(5) + log(3) rewrite the following in the form log(c).
Dennis_Churaev [7]
Log (5) + log (3) = log (15)
3 0
3 years ago
Read 2 more answers
Which composition of transformations maps figure EFGH to figure E"F"G"H"?
Lady bird [3.3K]

The transformations would map EFGH to {\text{E''F''G''H''}} is a reflection across line k followed by a translation down. Option (a) is correct.

Further explanation:

Given:

The compositions of transformations from EFGH to {\text{E''F''G''H''}} are as follows,

(a). A reflection across line k followed by a translation down.

(b). A translation down followed by a reflection across line k.

(c). A {180^ \circ } rotation about point G followed by a translation to the right.

(d). A translation to the right followed by a {180^ \circ } rotation about point G.

Explanation:

Translation can be defined as to move the function to a certain displacement. If the points of a line or any objects are moved in the same direction it is a translation.

Rotation is defined as a movement around its own axis. A circular movement is a rotation.

The transformations would map EFGH to {\text{E''F''G''H''}} is a reflection across line k followed by a translation down. Option (a) is correct.

Option (a) is correct.

Option (b) is not correct.

Option (c) is not correct.

Option (d) is not correct.

Learn more:

  1. Learn more about inverse of the functionhttps://brainly.com/question/1632445.
  2. Learn more about equation of circle brainly.com/question/1506955.
  3. Learn more about range and domain of the function brainly.com/question/3412497

Answer details:

Grade: Middle School

Subject: Mathematics

Chapter: Triangles

Keywords: rotation, transformation, map, EFGH, composition, translation, triangle, rotation about point A, mapped, triangle pair, mapping, equal angles, sides, congruent, two triangles, common point.

4 0
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Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE
EleoNora [17]

Answer:

0, 10

Step-by-step explanation:

The given function is:

g(y) = \frac{y-5}{y^2-3y+15}

According to the quotient rule:

d(\frac{f(y)}{h(y)})  = \frac{f(y)*h'(y)-h(y)*f'(y)}{h^2(y)}

Applying the quotient rule:

g(y) = \frac{y-5}{y^2-3y+15}\\g'(y)=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}

The values for which g'(y) are zero are the critical points:

g'(y)=0=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}\\(y-5)*(2y-3)-(y^2-3y+15)=0\\2y^2-3y-10y+15-y^2+3y-15\\y^2-10y=0\\y=\frac{10\pm \sqrt 100}{2}\\y_1=\frac{10-10}{2}= 0\\y_2=\frac{10+10}{2}=10

The critical values are y = 0 and y = 10.

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3 years ago
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